Cellulose is a polysaccharide composed of individual anhydroglucose units which are linked through a glycosidic bond (FIG. 16). The number ‘n’ of anhydroglucose units in the polymer chain is defined as the degree of polymerisation. Typically, production of cellulose ethers (CE's) involves replacing some of the hydroxyl hydrogen groups of cellulose with a substituent group, for example a methyl group, an ethyl group, a carboxymethyl group, a hydroxyehthyl group, a hydroxypropyl group, or some combination thereof. For example, a hydroxyethyl methyl cellulose (HEMC) may be produced by replacing some of the groups of cellulose with hydroxyethyl groups and methyl groups. Likewise, a hydroxypropyl methyl cellulose (HPMC) may be produced with hydroxypropyl and methyl groups replacing some of the hydroxyl groups of the cellulose.
The number of substituted hydroxyl groups per anhydroglucose unit is expressed as the degree of substitution (DS). The DS can vary between 0 and 3. As with all polymer reactions, this reaction does not occur uniformly along the polymer chain. The reported degree of substitution is therefore a mean degree of substitution over the whole polymer chain. Alternatively, molar substitution (MS) may be used to report the number of moles of substituent groups, such as a hydroxypropyl group, per mole of anhydroglucose. Often, manufacturers follow a convention whereby one of the substituents is reported by DS and the other by MS, where the substituent reported by MS may replace a hydroxyl group or may attach to another substituent to form a graft. The DS is not always reported, and the value reported is often inaccurate or given as a broad range, as shown in Table I.
In another alternative, the weight percent of substituents is reported. Weight percent of substituents may be directly related to DS and MS. For example, the following equations show the conversion for HPMC:                                           DS            ⁡                          (                              OCH                3                            )                                =                                                    wt                ⁢                                                                   ⁢                %                ⁢                                                                   ⁢                                  OCH                  3                                            31                        *                          162                              100                -                                  (                                                            wt                      ⁢                                                                                           ⁢                      %                      ⁢                                                                                           ⁢                                              OC                        3                                            ⁢                                              H                        6                                            ⁢                                              OH                        /                        1.29                                                              +                                          wt                      ⁢                                                                                           ⁢                      %                      ⁢                                                                                           ⁢                                              OCH                        3                                            *                      0.45                                                        )                                                                    ⁢                                  ⁢        and                            EQ        .                                   ⁢        1                                          MS          ⁡                      (                                          OC                3                            ⁢                              H                6                            ⁢              OH                        )                          =                                            wt              ⁢                                                           ⁢              %              ⁢                                                           ⁢                              OC                3                            ⁢                              H                6                            ⁢              OH                        75                    *                                    162                              100                -                                  (                                                            wt                      ⁢                                                                                           ⁢                      %                      ⁢                                                                                           ⁢                                              OC                        3                                            ⁢                                              H                        6                                            ⁢                                              OH                        /                        1.29                                                              +                                          wt                      ⁢                                                                                           ⁢                      %                      ⁢                                                                                           ⁢                                              OCH                        3                                            *                      0.45                                                        )                                                      .                                              EQ        .                                   ⁢        2            
Cellulose ethers are conventionally differentiated by type of substituent and the viscosity of an aqueous solution of the cellulose ether. For example methyl cellulose (MC), ethyl cellulose (EC), carboxymethyl cellulose (CMC), hydroxyethyl cellulose (HEC), ethyl hydroxyethyl cellulose (EHEC), ethyl hydroxypropyl cellulose (EHPC) and hydroxypropyl cellulose (HPC) are named for the type of substituent group used to replace the hydroxyl group in cellulose. The viscosity of an aqueous solution including a cellulose ether is an important characteristic for its typical use as a thickener; therefore, cellulose ethers are also differentiated by viscosity, which depends on the degree of polymerization (directly related to the measured molecular weight), and the type and degree of substitution of substituent groups. As the molecular weight increases, the viscosity of an aqueous solution of the cellulose ether increases also. However, the effect of the degree of substitution depends on the particular type of substituent group, which may also affect the viscosity of the cellulose ether.
Manufacturers characterize the effect of a particular cellulose ether on the viscosity by reporting the measured viscosity of a 2 wt % aqueous solution of the cellulose ether at 20° C. Herein, we refer to this 2 wt % viscosity as the viscosity grade of the particular cellulose ether. Typically, the viscosity grade is measured by one of two techniques: Brookfield and Ubbelohde. Often, the measured viscosity grade differs between the two techniques. For example, results using both techniques are shown in Table I for some cellulose ethers.
A modified cellulose, also referred to herein as a cellulose derivative, is used in plaster and joint compounds as a thickener (or to modify the rheology in some way) and to improve the workability of gypsum-based compounds. Cellulose ethers have been known to improve some other properties, including the consistency, adhesion and water retention of gypsum-based joint compounds and tile adhesives. However, adhesion, water retention and thickening are considered detrimental in the production of some gypsum-based composite structures. For example, wallboard is processed from a slurry that is continuously mixed and fed onto a belt. Thus, it is desirable for the slurry used to make wallboard to be more readily mixed and more quickly dried than a plaster, which is ordinarily formed or shaped by hand in a manual layering process.
Calcined gypsum powder (calcium sulfate hemihydrate and/or calcium sulfate anhydrite) is usually mixed with water and a variety of additives, which are used for nucleating the growth of gypsum crystals, as anti-fungal agents, light weight fillers or reinforcing fibers. Dissolution of the calcined gypsum powder in the water and a resulting hydration reaction causes crystallization of gypsum crystals (calcium sulfate dihydrate) forming a core of interlocking gypsum crystals. Application of multi-ply face sheets is usually integrated with the formation of the core into a wallboard. This is often followed by mild heating to drive off the remaining free (unreacted) water to yield a dry product, having face sheets adhered to gypsum core.
Tensile and shear strength of conventional wallboard is related primarily to the strength of the facing paper, typically an oriented fiber, multi-ply facing paper that is applied to the gypsum-based slurry during a continuous forming process. For a ½ inch wallboard with a density of about 0.6 g/cc, approximately one-half of the nail pull resistance and two-thirds of the flexural strength are supplied by the paper face sheets, which also account for 40% of the manufacturing costs. The core is usually exceptionally poor at handling tensile loads of any kind
Even compressive loads are limited for lower density wallboards without face sheets. Adding porosity and/or a low-density, expanded filler (e.g. perlite) into the conventional material reduces the core density. However, the strength of gypsum sheets decreases dramatically with density. For example, a dramatic decrease of the nail-pull resistance with density of ½-inch gypsum wallboard, both papered and non-papered, can be seen in FIG. 3.
Several unique challenges have thus far restricted the commercialization of polymer-reinforced cementitious products to relatively expensive niche products. For example, the nail pull resistance may decrease with the addition of some organic additives or an increase in nail pull resistance may require concentrations of polymers greater than 5 wt %, which can lead to problems such as inflammability, reduced extinguishability, commercially unacceptable cost of the wallboard, and mold susceptibility. Therefore, there is a longstanding and unresolved need for an additive that can increase both the nail pull resistance and the flexural strength of wallboard core, allowing the core density to be reduced, without adversely affecting other properties of the gypsum-based product.